Thermodynamics: Comprehensive NEET Chemistry Notes

1. Introduction to Thermodynamics

Thermodynamics is the study of energy transformations, dealing with macroscopic systems rather than microscopic systems. It involves understanding how energy is exchanged in chemical reactions and physical processes.

1.1 Basic Concepts

Thermodynamics focuses on systems, surroundings, and boundaries. A system is the part of the universe we study, and the surroundings are everything else. The universe is the sum of the system and its surroundings.

Did You Know?

The first law of thermodynamics is also known as the law of conservation of energy.

Real-life Application:

Thermodynamics principles are applied in designing engines and refrigerators.

2. Types of Systems

Systems in thermodynamics can be classified based on the exchange of energy and matter with the surroundings.

2.1 Open System

An open system exchanges both energy and matter with its surroundings. Example: A beaker of water exposed to the air.

2.2 Closed System

A closed system exchanges energy but not matter with its surroundings. Example: A sealed flask containing water.

2.3 Isolated System

An isolated system does not exchange energy or matter with its surroundings. Example: A thermos flask.

NEET Tip:

Remember the types of systems as they frequently appear in NEET questions.

3. Internal Energy

Internal energy (U) is the total energy contained within a system. It can change due to heat transfer or work done.

3.1 Work and Heat

Work (w) and heat (q) are the two ways to change the internal energy of a system.

Work (w): Energy transferred when an object is moved by a force.
Heat (q): Energy transferred due to temperature difference.

Example:

When you heat water, the energy is transferred as heat, increasing the water's internal energy.

NEET Problem-Solving Strategy:

Use the formula $Δ U = q + w$ to calculate changes in internal energy.

4. The First Law of Thermodynamics

The first law of thermodynamics states that energy cannot be created or destroyed, only transformed.

4.1 Mathematical Expression

The first law can be expressed as: $Δ U = q + w$ Where:

$Δ U$ = change in internal energy
$q$ = heat added to the system
$w$ = work done on the system

Mnemonic:

"Energy can't be created or destroyed, just moved around" to remember the essence of the first law.

5. Enthalpy (H)

Enthalpy is the heat content of a system at constant pressure. It is a state function and is given by: $H = U + P V$

5.1 Enthalpy Change

The change in enthalpy ( $Δ H$ ) for a reaction can be written as: $Δ H = Δ U + P Δ V$

Real-life Application:

Enthalpy changes are crucial in understanding energy requirements for chemical processes in industries.

6. Heat Capacity

Heat capacity is the amount of heat required to raise the temperature of a substance by one degree Celsius.

6.1 Molar and Specific Heat Capacity

Molar Heat Capacity (Cm): Heat capacity per mole of a substance.
Specific Heat Capacity (c): Heat capacity per unit mass of a substance.

Example:

Water has a high specific heat capacity, which is why it takes a long time to heat up.

Common Misconception:

Specific heat capacity is not the same as heat capacity. The former is per unit mass, while the latter is for the whole substance.

7. Thermochemical Equations

Thermochemical equations show the enthalpy changes during chemical reactions.

7.1 Standard Enthalpy of Formation

The standard enthalpy of formation ( $Δ_{f} H^{0}$ ) is the change in enthalpy when one mole of a compound is formed from its elements in their standard states.

Example:

The standard enthalpy of formation of water: $2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l); Δ_{f} H^{0} = - 285.8 kJ/mol$

NEET Tip:

Familiarize yourself with common thermochemical equations and their enthalpy changes.

8. Hess’s Law

Hess's Law states that the total enthalpy change of a reaction is the sum of the enthalpy changes of individual steps.

8.1 Application of Hess’s Law

Hess’s Law can be used to calculate enthalpy changes for reactions that are difficult to measure directly.

Example:

To find the enthalpy change for the formation of CO from graphite and oxygen, combine the enthalpy changes of two known reactions.

NEET Problem-Solving Strategy:

Use Hess’s Law for indirect calculations of enthalpy changes in complex reactions.

9. Spontaneity and Gibbs Energy

The spontaneity of a process is determined by Gibbs free energy ( $Δ G$ ).

9.1 Gibbs Free Energy Equation

$Δ G = Δ H - T Δ S$ Where:

$Δ G$ = Gibbs free energy change
$Δ H$ = enthalpy change
$Δ S$ = entropy change
$T$ = temperature in Kelvin

Did You Know?

A negative $Δ G$ indicates a spontaneous process.

Concept Connection:

Gibbs free energy connects thermodynamics with chemical kinetics and equilibrium.

Quick Recap

Thermodynamics deals with energy transformations in chemical reactions.
Systems can be open, closed, or isolated.
The first law of thermodynamics states that energy is conserved.
Enthalpy is the heat content of a system at constant pressure.
Heat capacity is the amount of heat required to change the temperature of a substance.
Hess's Law helps calculate enthalpy changes for complex reactions.
Gibbs free energy determines the spontaneity of a process.

Practice Questions

Question: Define the first law of thermodynamics. Solution: The first law of thermodynamics states that energy cannot be created or destroyed, only transformed.
Question: What is an isolated system? Give an example. Solution: An isolated system does not exchange energy or matter with its surroundings. Example: A thermos flask.
Question: Calculate the change in internal energy if 200 J of heat is added to a system and 50 J of work is done on it. Solution: $Δ U = q + w = 200 J + 50 J = 250 J$
Question: Explain the significance of Gibbs free energy in chemical reactions. Solution: Gibbs free energy determines whether a reaction is spontaneous. A negative $Δ G$ indicates a spontaneous reaction.
Question: State Hess’s Law and give its importance. Solution: Hess's Law states that the total enthalpy change of a reaction is the sum of the enthalpy changes of individual steps. It is important for calculating enthalpy changes indirectly.
Question: What is the enthalpy change for the reaction: $C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l) i f Δ_{f} H^{0} f or C H_{4} (g), C O_{2} (g), an d H_{2} O (l) a re - 74.8, - 393.5, an d - 285.8 k J / m o l$ respectively? Solution: $Δ_{r} H^{0} = [(- 393.5) + 2 (- 285.8)] - [- 74.8 + 2 (0)] = - 890.3 kJ/mol$
Question: Describe an adiabatic process. Solution: An adiabatic process is one in which no heat is exchanged with the surroundings.
Question: What is the relationship between heat capacity at constant pressure (Cp) and constant volume (Cv) for an ideal gas? Solution: $C_{p} - C_{v} = R$ , where $R$ is the universal gas constant.
Question: Why is water's high specific heat capacity important for the environment? Solution: Water's high specific heat capacity helps regulate Earth's temperature by absorbing and releasing large amounts of heat with little temperature change.
Question: Calculate the work done when a gas expands from 2 L to 5 L against a constant external pressure of 1 atm.

Solution:

$w = - P Δ V = - 1 atm \times (5 - 2) L = - 3 L atm$

$1 L atm = 101.3 J$

$w = - 3 \times 101.3 = - 303.9 J$

Quick Reference Guide and Glossary

System: Part of the universe under study.
Surroundings: Everything outside the system.
First Law of Thermodynamics: Energy cannot be created or destroyed.
Enthalpy (H): Heat content at constant pressure.
Heat Capacity: Amount of heat required to change temperature.
Hess's Law: Total enthalpy change is the sum of individual steps.
Gibbs Free Energy (G): Determines spontaneity of a process.